How to Determine Correct Inverse Trig Angle?

AI Thread Summary
Inverse trigonometric functions have a principal domain that restricts their outputs to a single value, which is why calculators only provide one answer. For example, arcsin(0.5) yields pi/6, as the arcsine function's domain is limited to [-pi/2, pi/2]. To find other angles that yield the same sine value, one can use the property that sin(pi - x = sin(x) and consider periodicity with sin(x + 2pi) = sin(x). The discussion emphasizes the importance of understanding these domains and properties to determine additional angles when needed. This foundational knowledge is crucial for tackling more complex problems in trigonometry.
Ascendant0
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Homework Statement
Not a specific homework problem, but more so that the book indicates that for inverse trig functions sin, cos, and tan, there's usually another possible answer that the calculator won't give. But, it doesn't explain *how* to find the other angle it doesn't give?
Relevant Equations
Just the sin, cos, and tan inverse functions
I understand why certain inverse trig functions have two answers. Like for arcsin(0.5), it could be pi/6 or 5pi/6. I know both angles have the same sin value, that they're both on the same horizontal line on a graph of sin, I get all of that, but two questions about it:

1) In cases where it's not a special trig angle that I can refer to on the chart, how do I determine what the other possible answer could be?

2) Even after I know how to do that, the book simply says to "determine the more reasonable one for the given situation," which sure, I can do for now for these easier problems, but when there are far more advanced problems in the future, where determining the "more reasonable" angle isn't viable, in those instances, how do you determine which is the correct angle?
 
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Ascendant0 said:
Homework Statement: Not a specific homework problem, but more so that the book indicates that for inverse trig functions sin, cos, and tan, there's usually another possible answer that the calculator won't give. But, it doesn't explain *how* to find the other angle it doesn't give?

I understand why certain inverse trig functions have two answers. Like for arcsin(0.5), it could be pi/6 or 5pi/6.
The inverse trig functions each have ***one*** answer, otherwise they would be merely relations, not functions. If I use a calculator to approximate ##\sin^{-1}(0.5)## I get the decimal equivalent of ##\frac \pi 6##.
What you're missing is that each of the inverse trig functions has a principal domain. For the arcsine function, the principal domain is the interval ##[-\frac \pi 2, \frac \pi 2]##; for the arccosine function, the principal domain is ##[0, \pi]##. The domain for the arctangent function is the same as for the arcsine function, but with the endpoints removed.

To get other values for which, for example, ##\sin(x) = 0.5## look at the graph of the sine function and notice that ##\sin(\pi - x) = \sin(x)## combined with the fact that ##\sin(x + 2\pi) - \sin(x)##. And similar for the other trig functions and inverses.
 
Mark44 said:
The inverse trig functions each have ***one*** answer, otherwise they would be merely relations, not functions. If I use a calculator to approximate ##\sin^{-1}(0.5)## I get the decimal equivalent of ##\frac \pi 6##.
What you're missing is that each of the inverse trig functions has a principal domain. For the arcsine function, the principal domain is the interval ##[-\frac \pi 2, \frac \pi 2]##; for the arccosine function, the principal domain is ##[0, \pi]##. The domain for the arctangent function is the same as for the arcsine function, but with the endpoints removed.

To get other values for which, for example, ##\sin(x) = 0.5## look at the graph of the sine function and notice that ##\sin(\pi - x) = \sin(x)## combined with the fact that ##\sin(x + 2\pi) - \sin(x)##. And similar for the other trig functions and inverses.
Thank you for that clarification. Wrote that down for future reference.

And you can see here in this screenshot of the book what I was talking about:

Wrong it seems.JPG
 
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